Question: Simplify and expand the following expression: $ \dfrac{2}{3x + 24}+ \dfrac{2}{3x + 3}- \dfrac{4}{x^2 + 9x + 8} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3x + 24} = \dfrac{2}{3(x + 8)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{2}{3x + 3} = \dfrac{2}{3(x + 1)}$ We can factor the quadratic in the third term: $ \dfrac{4}{x^2 + 9x + 8} = \dfrac{4}{(x + 8)(x + 1)}$ Now we have: $ \dfrac{2}{3(x + 8)}+ \dfrac{2}{3(x + 1)}- \dfrac{4}{(x + 8)(x + 1)} $ The least common multiple of the denominators is: $ 9(x + 8)(x + 1)$ In order to get the first term over $9(x + 8)(x + 1)$ , multiply by $\dfrac{3(x + 1)}{3(x + 1)}$ $ \dfrac{2}{3(x + 8)} \times \dfrac{3(x + 1)}{3(x + 1)} = \dfrac{6(x + 1)}{9(x + 8)(x + 1)} $ In order to get the second term over $9(x + 8)(x + 1)$ , multiply by $\dfrac{3(x + 8)}{3(x + 8)}$ $ \dfrac{2}{3(x + 1)} \times \dfrac{3(x + 8)}{3(x + 8)} = \dfrac{6(x + 8)}{9(x + 8)(x + 1)} $ In order to get the third term over $9(x + 8)(x + 1)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{4}{(x + 8)(x + 1)} \times \dfrac{9}{9} = \dfrac{36}{9(x + 8)(x + 1)} $ Now we have: $ \dfrac{6(x + 1)}{9(x + 8)(x + 1)} + \dfrac{6(x + 8)}{9(x + 8)(x + 1)} - \dfrac{36}{9(x + 8)(x + 1)} $ $ = \dfrac{ 6(x + 1) + 6(x + 8) - 36} {9(x + 8)(x + 1)} $ Expand: $ = \dfrac{6x + 6 + 6x + 48 - 36}{9x^2 + 81x + 72} $ $ = \dfrac{12x + 18}{9x^2 + 81x + 72}$ Simplify: $ = \dfrac{4x + 6}{3x^2 + 27x + 24}$